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In triangle $ABC,$   $AP = BQ = \frac{PQ}{2}$ and $CR/RA = 3/2.$  Find the ratio of the area of triangle $BQC$  to the area of triangle $CRQ$

[asy]
pair A, X, Y, B, Z, C;
A = (0,0);
X = (0.3,-0.2);
Z = (0.4,0.2);
Y = 3*X;
B = 4*X;
C = 2.5*Z;
draw(X--Z--Y--C--B--A--C);
label("$A$",A,W);
label("$B$",B,S);
label("$C$",C,N);
label("$P$",X,SW);
label("$Q$",Y,SW);
label("$R$",Z,NW);
[/asy]

Please Try to Respond as Fast as possible.

 Feb 12, 2020
 #1
avatar+2850 
+3

In triangle \(ABC\)\(AP = BQ = \frac{PQ}{2}\) and \(CR/RA = 3/2 \). Find the ratio of the area of triangle \(BQC\)  to the area of triangle \(CRQ\).

Since AP = BQ = PQ/2. We can conclude that QB is 1/4 of AB.

 

Let the area of ABC = x

 

So BQC = \(\frac{x}{4}\)

 

Area of triangle CAQ = \(x-\frac{x}{4}=\frac{3x}{4}\)

 

Since CR/RA = 3/2, that means that CRQ = \(\frac{3x}{4}*\frac{3}{5}=\frac{9x}{20}\)

 

 

SO

Area of BQC is \(\frac{x}{4}\)

Area of CRQ is \(\frac{9x}{20}\)

 


Can you find the ratio now? (Hint: It's simple algebra!)

 Feb 12, 2020

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